Question

I need a thorough explanation on how you got your answer and the graphing calculator/equation you used to solve it.

Answer 1

We will need to use the following formulas

`\begin{gathered} X=(1)/(N)\sum ^N_(i=1)x_i \\ Y=(1)/(N)\sum ^N_(i=1)y_i \\ m=(\sum^N_(i=1)(x_i-X)(y_i-Y))/(\sum^N_(i=1)(x_i-X)^2) \\ \text{and} \\ r=\frac{\sum^N_(i=1)(x_i-X)(y_i-Y)}{\sqrt{\sum^N_(i=1)(x_i-X)^2\sum^N_(i=1)(y_i_{}-Y)}^2^{}} \\ \text{and} \\ Y=mX+b \end{gathered}`

In our case, using a calculator

`X=5.5,Y=555.6\to\text{ means of the values of x and y, respectively}`

Therefore,

`\begin{gathered} SS_x=\sum ^N_(i=1)(x_i-X)^2=82.5 \\ SP=\sum ^N_(i=1)(x_i-X)(y_i-Y)=1153 \\ \Rightarrow m=(1153)/(82.5)=13.975\ldots \\ \Rightarrow m\approx14 \end{gathered}`

Finding b,

`\begin{gathered} Y=mX+b \\ \Rightarrow b=Y-mX \\ \Rightarrow b=555.6-(1153)/(82.5)\cdot5.5=478.73\ldots \\ \Rightarrow b\approx479 \end{gathered}`

Thus, the equation of the line of best is y=14x+479.

As for the correlation coefficient,

`\begin{gathered} \sum ^N_(i=1)(x_i-X)(y_i-Y)=1153 \\ \sum ^N_(i=1)(x_i-X)^2=SS_x=82.5 \\ and \\ \sum ^N_(i=1)(y_i-Y)^2=SS_y=16118.4 \end{gathered}`

Therefore,

`\begin{gathered} \Rightarrow r=\frac{1153}{\sqrt[]{82.5\cdot16118.4}}=0.9998\ldots\approx1 \\ \Rightarrow r=1 \end{gathered}`

Then, the Correlation Coefficient is r=1.

The r is practically 1, the model is considered a good fit; in fact, it is a perfect fit (strong relationship)

Summation Notation

Consider the expression

`\sum ^N_(i=1)x_i`

Notice that there are 10 values of x in the table, order them in the following way

`x_1=1,x_2=2,\ldots,x_(10)=10`

Then, after setting N=10,

`\sum ^(10)_(i=1)x_i=x_1+x_2+\cdots+x_(10)=1+2+\ldots+10=55`

Another example,

`\begin{gathered} \sum ^N_(i=1)y_i,N=10 \\ \Rightarrow\sum ^(10)_(i=1)y_i=y_1+y_2+\cdots+y_(10)=492+507+\cdots+618 \end{gathered}`

We found that X=5.5 (see above); then,